post_href stringlengths 57 213 | python_solutions stringlengths 71 22.3k | slug stringlengths 3 77 | post_title stringlengths 1 100 | user stringlengths 3 29 | upvotes int64 -20 1.2k | views int64 0 60.9k | problem_title stringlengths 3 77 | number int64 1 2.48k | acceptance float64 0.14 0.91 | difficulty stringclasses 3
values | __index_level_0__ int64 0 34k |
|---|---|---|---|---|---|---|---|---|---|---|---|
https://leetcode.com/problems/number-of-dice-rolls-with-target-sum/discuss/2650329/Python-or-DP-or-95-Space-Optimized-Solution-or | class Solution:
def numRollsToTarget(self, n: int, k: int, target: int) -> int:
module = 10**9 + 7
front = [0]*(target+1)
# base case..
for tar in range(target+1):
if tar >= 1 and tar <= k:
front[tar] = 1
else:
front[tar] = 0
... | number-of-dice-rolls-with-target-sum | Python | DP | 95% Space Optimized Solution | | quarnstric_ | 0 | 7 | number of dice rolls with target sum | 1,155 | 0.536 | Medium | 17,900 |
https://leetcode.com/problems/number-of-dice-rolls-with-target-sum/discuss/2650177/Python-simple-recursion-with-memoization | class Solution:
def numRollsToTarget(self, d: int, f: int, target: int) -> int:
mod=(10**9)+7
memo={}
def tryforother(d,f,target,memo):
if target<d or target>d*f:
return 0
if d==1:
return 1 if target<=f else 0
if (d,f,target... | number-of-dice-rolls-with-target-sum | Python simple recursion with memoization | iliyazali | 0 | 2 | number of dice rolls with target sum | 1,155 | 0.536 | Medium | 17,901 |
https://leetcode.com/problems/number-of-dice-rolls-with-target-sum/discuss/2650115/Easy-explained-Dynamic-Programming-solution-on-Python | class Solution:
def numRollsToTarget(self, n: int, k: int, target: int) -> int:
limit = 10**9 + 7
dp = [0 for x in range(target + 1)]
dp[0] = 1
for roll_number in range(n):
for sum_points in reversed(range(target + 1)):
if sum_points < roll_number:
... | number-of-dice-rolls-with-target-sum | Easy explained Dynamic Programming solution on Python | kisel_dv | 0 | 16 | number of dice rolls with target sum | 1,155 | 0.536 | Medium | 17,902 |
https://leetcode.com/problems/number-of-dice-rolls-with-target-sum/discuss/2650053/Number-of-Dice-Rolls-With-Target-Sum-Python-or-Java | class Solution:
def numRollsToTarget(self, d: int, f: int, target: int) -> int:
memo = {}
def dp(d, target):
if d == 0:
return 0 if target > 0 else 1
if (d, target) in memo:
return memo[(d, target)]
to_return = 0
for k i... | number-of-dice-rolls-with-target-sum | Number of Dice Rolls With Target Sum [ Python | Java ] | klu_2100031497 | 0 | 68 | number of dice rolls with target sum | 1,155 | 0.536 | Medium | 17,903 |
https://leetcode.com/problems/number-of-dice-rolls-with-target-sum/discuss/2650012/Python-simple-DP-solution | class Solution:
def numRollsToTarget(self, n: int, k: int, target: int) -> int:
dp = [1] + [0 for _ in range(target)]
for _ in range(n):
n_dp = [0 for _ in range(target+1)]
for i in range(target+1):
if dp[i] != 0:
for j in range(1,k+1):
... | number-of-dice-rolls-with-target-sum | Python simple DP solution | AllenXia | 0 | 3 | number of dice rolls with target sum | 1,155 | 0.536 | Medium | 17,904 |
https://leetcode.com/problems/number-of-dice-rolls-with-target-sum/discuss/2649360/Python-Simple-Python-Solution | class Solution:
def numRollsToTarget(self, n: int, k: int, target: int) -> int:
dp = [[-1 for j in range(target+2)] for i in range(n+1)]
def cal(rem_d, tot_S):
if rem_d == 0 and tot_S == 0:
return 1
if tot_S < 0 or rem_d <= 0:
... | number-of-dice-rolls-with-target-sum | [ Python ] ✅ Simple Python Solution ✅✅ | vaibhav0077 | 0 | 123 | number of dice rolls with target sum | 1,155 | 0.536 | Medium | 17,905 |
https://leetcode.com/problems/number-of-dice-rolls-with-target-sum/discuss/2649360/Python-Simple-Python-Solution | class Solution:
def numRollsToTarget(self, n: int, k: int, target: int) -> int:
dp = [[0 for j in range(target+2)] for i in range(n+1)]
dp[0][0] = 1
for rem_d in range(1, n+1):
for tot_S in range(0, target + 2):
for a in range(1,min(k+1,tot_S + 1)):
... | number-of-dice-rolls-with-target-sum | [ Python ] ✅ Simple Python Solution ✅✅ | vaibhav0077 | 0 | 123 | number of dice rolls with target sum | 1,155 | 0.536 | Medium | 17,906 |
https://leetcode.com/problems/number-of-dice-rolls-with-target-sum/discuss/2649248/2D-Dynamic-programming | class Solution:
def numRollsToTarget(self, n: int, k: int, target: int) -> int:
dp = [[0 for _ in range(target+1)] for _ in range(n+1)]
dp[0][0] = 1
for t in range(target+1):
for i in range(n):
for face in range(1,k+1):
if t-face >= 0:
... | number-of-dice-rolls-with-target-sum | 2D Dynamic programming | chris1nexus | 0 | 6 | number of dice rolls with target sum | 1,155 | 0.536 | Medium | 17,907 |
https://leetcode.com/problems/number-of-dice-rolls-with-target-sum/discuss/2649168/Python3%3A-Faster-than-99.35 | class Solution:
def numRollsToTarget(self, n: int, k: int, target: int) -> int:
if target < n or target > k * n:
return 0
MAX = 10 ** 9 + 7
def binom(x, y):
if y > x or x < 0:
return 0
if y == 0 or y == x:
return 1
... | number-of-dice-rolls-with-target-sum | Python3: Faster than 99.35% | Odinnnnnn | 0 | 75 | number of dice rolls with target sum | 1,155 | 0.536 | Medium | 17,908 |
https://leetcode.com/problems/number-of-dice-rolls-with-target-sum/discuss/2649136/Python-or-Triple-loop-DP | class Solution:
def numRollsToTarget(self, n: int, k: int, target: int) -> int:
modd=10**9+7
dp=[[0 for i in range(target+1)]for j in range(n+1)]
for i in range(1,min(target+1,k+1)):
dp[1][i]=1
for num in range(2,n+1):
for s... | number-of-dice-rolls-with-target-sum | Python | Triple loop DP | Prithiviraj1927 | 0 | 47 | number of dice rolls with target sum | 1,155 | 0.536 | Medium | 17,909 |
https://leetcode.com/problems/number-of-dice-rolls-with-target-sum/discuss/2648407/Python-solution-via-backtracking-and-cache-faster-than-89 | class Solution:
from functools import cache
@cache
def numRollsToTarget(self, n: int, k: int, target: int) -> int:
if target > n * k:
return 0
elif target < 0 or (target == 0 and n > 0):
return 0
elif target == 0 and n == 0:
return 1
else... | number-of-dice-rolls-with-target-sum | Python solution via backtracking and cache faster than 89% | Terry_Lah | 0 | 6 | number of dice rolls with target sum | 1,155 | 0.536 | Medium | 17,910 |
https://leetcode.com/problems/number-of-dice-rolls-with-target-sum/discuss/2648347/Python-Bottom-up-DP | class Solution:
def numRollsToTarget(self, n: int, k: int, target: int) -> int:
MOD = 10 ** 9 + 7
# DP (n, target)
dp = [[0] * (target+1+k+1+1) for _ in range(n+1)]
for i in range(1, k+1):
dp[1][i] = 1
for i in range(1, n):
for t in range(0, target+1... | number-of-dice-rolls-with-target-sum | [Python] Bottom-up DP | wtain | 0 | 5 | number of dice rolls with target sum | 1,155 | 0.536 | Medium | 17,911 |
https://leetcode.com/problems/number-of-dice-rolls-with-target-sum/discuss/2648347/Python-Bottom-up-DP | class Solution:
def numRollsToTarget(self, n: int, k: int, target: int) -> int:
MOD = 10 ** 9 + 7
# DP (n, target)
dp = [0] * (target+k+1)
for i in range(1, k+1):
dp[i] = 1
for i in range(n-1):
next_dp = [0] * (target+k+1)
for t in range(... | number-of-dice-rolls-with-target-sum | [Python] Bottom-up DP | wtain | 0 | 5 | number of dice rolls with target sum | 1,155 | 0.536 | Medium | 17,912 |
https://leetcode.com/problems/number-of-dice-rolls-with-target-sum/discuss/2250067/PYTHON-SOL-or-MEMO-%2B-RECURSION-or-WELL-EXPLAINED-or-EASY-or-FAST-or | class Solution:
def recursion(self,n,k,target):
if n == 1:
# base case
return 1 if 1 <= target <= k else 0
if (n,target) in self.dp: return self.dp[(n,target)]
ans = 0
for i in range(1,k+1):
ans += self.recursion(n-1,k,target - i)
self.dp[(... | number-of-dice-rolls-with-target-sum | PYTHON SOL | MEMO + RECURSION | WELL EXPLAINED | EASY | FAST | | reaper_27 | 0 | 134 | number of dice rolls with target sum | 1,155 | 0.536 | Medium | 17,913 |
https://leetcode.com/problems/number-of-dice-rolls-with-target-sum/discuss/2221149/Python-bottom-up-DP | class Solution:
def numRollsToTarget(self, n: int, k: int, target: int) -> int:
if n == 1:
if target > k: return 0
else: return 1
memo = [0 for t in range(target+1)]
# Note that we do not use memo[0]
for i in range(1, target+1):
if i <= k:
... | number-of-dice-rolls-with-target-sum | Python bottom-up DP | sticky_bits | 0 | 113 | number of dice rolls with target sum | 1,155 | 0.536 | Medium | 17,914 |
https://leetcode.com/problems/swap-for-longest-repeated-character-substring/discuss/2255000/PYTHON-or-AS-INTERVIEWER-WANTS-or-WITHOUT-ITERTOOLS-or-WELL-EXPLAINED-or | class Solution:
def maxRepOpt1(self, text: str) -> int:
first_occurence,last_occurence = {},{}
ans,prev,count = 1,0,0
n = len(text)
for i in range(n):
if text[i] not in first_occurence: first_occurence[text[i]] = i
last_occurence[text[i]] = i
... | swap-for-longest-repeated-character-substring | PYTHON | AS INTERVIEWER WANTS | WITHOUT ITERTOOLS | WELL EXPLAINED | | reaper_27 | 1 | 178 | swap for longest repeated character substring | 1,156 | 0.454 | Medium | 17,915 |
https://leetcode.com/problems/swap-for-longest-repeated-character-substring/discuss/2825889/Python-Sliding-Window-O(N) | class Solution:
def maxRepOpt1(self, text: str) -> int:
freq = {}
for ch in text:
freq[ch] = freq.get(ch, 0)+1
n=len(text)
i=0
distinct = 0
cur={}
def include(ind):
nonlocal distinct
ch = text[ind]
cur[c... | swap-for-longest-repeated-character-substring | Python Sliding Window O(N) | saijayavinoth | 0 | 2 | swap for longest repeated character substring | 1,156 | 0.454 | Medium | 17,916 |
https://leetcode.com/problems/swap-for-longest-repeated-character-substring/discuss/1490882/Python-O(n)-solution | class Solution:
def maxRepOpt1(self, text: str) -> int:
intervals = collections.defaultdict(list)
prev = ''
start = 0
# collect intervals for each letter
for i, ch in enumerate(text):
if ch != prev and i != 0:
intervals[prev].append((start, i))
... | swap-for-longest-repeated-character-substring | Python O(n) solution | arsamigullin | 0 | 359 | swap for longest repeated character substring | 1,156 | 0.454 | Medium | 17,917 |
https://leetcode.com/problems/swap-for-longest-repeated-character-substring/discuss/1480970/Groupby-and-Counter | class Solution:
def maxRepOpt1(self, text: str) -> int:
lst = [(key, len(list(seq))) for key, seq in groupby(text)]
len_lst_2 = len(lst) - 2
cnt = Counter(text)
max_len = 0
for i, (key, n) in enumerate(lst):
max_len = max(max_len, n)
if n < cnt[key]:
... | swap-for-longest-repeated-character-substring | Groupby and Counter | EvgenySH | 0 | 99 | swap for longest repeated character substring | 1,156 | 0.454 | Medium | 17,918 |
https://leetcode.com/problems/swap-for-longest-repeated-character-substring/discuss/816103/Python-O(n)-solution-with-comments | class Solution:
def maxRepOpt1(self, text: str) -> int:
inuse = collections.defaultdict(int) # chars used in the repeated substring
left = collections.defaultdict(int) # not used chars
MOVE_TO_THE_NEXT_CHAR = 1
REPLACE = 0
LEAVE_AS_IT_IS = -1
res, i, n = 1, 0, 0
... | swap-for-longest-repeated-character-substring | Python O(n) solution with comments | arsamigullin | 0 | 260 | swap for longest repeated character substring | 1,156 | 0.454 | Medium | 17,919 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/2177578/Python3-O(n2)-oror-O(1)-Runtime%3A-96ms-97.20-Memory%3A-14.5mb-84.92 | class Solution:
# O(n^2) || O(1)
# Runtime: 96ms 97.20% Memory: 14.5mb 84.92%
def countCharacters(self, words: List[str], chars: str) -> int:
ans=0
for word in words:
for ch in word:
if word.count(ch)>chars.count(ch):
break
else:
... | find-words-that-can-be-formed-by-characters | Python3 O(n^2) || O(1) Runtime: 96ms 97.20% Memory: 14.5mb 84.92% | arshergon | 5 | 350 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,920 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/2188726/Python-1-liner | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
return sum(len(word) if collections.Counter(word) <= collections.Counter(chars) else 0 for word in words) | find-words-that-can-be-formed-by-characters | Python 1-liner | russellizadi | 2 | 205 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,921 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/2099811/Python-top-95-solution | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
ans = ''
for word in words:
for letter in word:
if chars.count(letter) < word.count(letter):
break
else:
ans += word
return len(ans) | find-words-that-can-be-formed-by-characters | Python top 95% solution | StikS32 | 2 | 253 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,922 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/1875631/Python-(Simple-Approach-and-Beginner-Friendly) | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
output = 0
for i in words:
count = 0
for j in i:
if chars.count(j) >= i.count(j):
count+=1
else:
break
if count ... | find-words-that-can-be-formed-by-characters | Python (Simple Approach and Beginner-Friendly) | vishvavariya | 2 | 160 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,923 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/1918471/Python-One-Liner-or-Counter | class Solution:
def countCharacters(self, words, chars):
d, total = Counter(chars), 0
for w in words: total += self.helper(w, d.copy())
return total
def helper(self, w, d):
for c in w:
if c not in d or d[c] == 0: return 0
else: d[c]-=1
return ... | find-words-that-can-be-formed-by-characters | Python - One-Liner | Counter | domthedeveloper | 1 | 82 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,924 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/1918471/Python-One-Liner-or-Counter | class Solution:
def countCharacters(self, words, chars):
return (lambda c:sum(len(x) for x in words if Counter(x) < c))(Counter(chars)) | find-words-that-can-be-formed-by-characters | Python - One-Liner | Counter | domthedeveloper | 1 | 82 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,925 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/1806893/Python3-Solution | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
c = list(chars)
l,ans = 0,0
for i in words:
for j in list(i):
if j in c:
l += 1
c.remove(j)
if l == len(i):
ans += l... | find-words-that-can-be-formed-by-characters | ✔Python3 Solution | Coding_Tan3 | 1 | 174 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,926 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/1386031/Python3-Faster-Than-99.94-Memory-Less-Than-79.96 | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
from collections import Counter
c = Counter(chars)
cnt = 0
for word in words:
good = True
for letter in word:
if word.count(letter) > c[letter]:
... | find-words-that-can-be-formed-by-characters | Python3 Faster Than 99.94%, Memory Less Than 79.96% | Hejita | 1 | 109 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,927 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/1177497/python-sol-faster-than-96-less-mem-than-93 | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
tot = 0
for i in range(len(words)):
for j in range(len(words[i])):
if (words[i][j] not in chars):
break
if (words[i].count(words[i][j]) > chars.count(words[... | find-words-that-can-be-formed-by-characters | python sol faster than 96% , less mem than 93% | elayan | 1 | 472 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,928 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/2822917/Python-oror-Faster-than-99.85-and-Memory-beats-85.14 | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
c = 0
for word in words:
good = True
for l in word:
if chars.count(l) < word.count(l):
good = False
break
if good == True: c +... | find-words-that-can-be-formed-by-characters | Python ✅✅✅|| Faster than 99.85% and Memory beats 85.14% | qiy2019 | 0 | 5 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,929 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/2813313/Simple-Python3-Solution-Runtime-101-ms-Beats-97.99-Memory-14.5-MB-Beats-85.3 | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
result_words = []
for word in words:
found = True
for ch in word:
if (ch not in chars) or (word.count(ch) > chars.count(ch)):
found = False
... | find-words-that-can-be-formed-by-characters | Simple Python3 Solution Runtime 101 ms Beats 97.99% Memory 14.5 MB Beats 85.3% | SupriyaArali | 0 | 4 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,930 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/2724771/Easy-Solution%3A-99.22-Faster | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
op=[]
for i in words:
na = 0
for j in i:
if (j not in chars)or (i.count(j)>chars.count(j)):
na = 1
break
if na==0:
... | find-words-that-can-be-formed-by-characters | Easy Solution: 99.22% Faster | BAparna97 | 0 | 9 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,931 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/2673811/Python-solution-easy-to-understand | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
c = Counter(chars)
ans = 0
for w in words:
temp = Counter(w)
found = True
for k, v in temp.items():
if k in c:
if c[k] < v:
... | find-words-that-can-be-formed-by-characters | Python solution - easy to understand | phantran197 | 0 | 11 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,932 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/2660999/Python%2BCounter | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
return sum([len(word) for word in words if Counter(word) <=Counter(chars) ]) | find-words-that-can-be-formed-by-characters | Python+Counter | Leox2022 | 0 | 3 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,933 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/2602478/Python-1-liner | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
return sum(len(word) for word in words if Counter(word) <= Counter(chars)) | find-words-that-can-be-formed-by-characters | Python 1-liner | Potentis | 0 | 41 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,934 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/2599299/91-ms-faster-than-99.21-(with-explanation) | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
charMap, result = [0] * 26, 0
for char in chars:
charMap[ord(char) - 97] += 1
for word in words:
if len(word) > len(chars):
continue
temp = charMap[::]
... | find-words-that-can-be-formed-by-characters | 91 ms, faster than 99.21% (with explanation) | kcstar | 0 | 35 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,935 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/2592278/A-Double-Dictionary-Approach | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
charMap, result = {}, 0
for char in chars:
charMap[char] = charMap.get(char, 0) + 1
for word in words:
if len(word) > len(chars):
continue
temp, count = {}, 0
... | find-words-that-can-be-formed-by-characters | A Double Dictionary Approach | kcstar | 0 | 14 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,936 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/2508114/easy-Python-Solution-92-faster | class Solution(object):
def countCharacters(self, words, chars):
x=set(chars)
ans=0
for i in words:
a=set(i)
if a.issubset(x):
arr=[o for o in a if chars.count(o)<i.count(o)]
if len(arr)==0:
ans+=len(i)
retur... | find-words-that-can-be-formed-by-characters | easy Python Solution 92% faster | pranjalmishra334 | 0 | 53 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,937 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/2395057/Easy-hashmap | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
d1={}
for i in chars:
if i in d1:
d1[i]+=1
else:
d1[i]=1
c=0
for i in words:
yes=len(i)
d={}
for j in i:
... | find-words-that-can-be-formed-by-characters | Easy hashmap | sunakshi132 | 0 | 75 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,938 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/2294765/Python3-Simple-Solution | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
count = 0
d = {}
dd = {}
for c in chars:
d[c] = d.get(c, 0) + 1
dd[c] = dd.get(c, 0) + 1
for word in words:
flag = True
for w in w... | find-words-that-can-be-formed-by-characters | Python3 Simple Solution | mediocre-coder | 0 | 131 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,939 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/2175654/2-Easy-Methods | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
hmap1=Counter(chars)
res=0
for word in words:
if not (Counter(word) - hmap1):
res+=len(word)
return res | find-words-that-can-be-formed-by-characters | 2 Easy Methods | Defence | 0 | 42 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,940 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/2067337/python3-easy-solution | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
l1=[]
sum=0
for i in range(len(words)):
temp=list(words[i])
l1=list(chars)
# print(l1)
c=0
for i in range(len(temp)):
if temp[i] in l1:... | find-words-that-can-be-formed-by-characters | python3 easy solution | vishwahiren16 | 0 | 90 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,941 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/2043636/python-3-oror-counter-solution-oror-O(C)O(1) | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
charsCount = collections.Counter(chars)
res = 0
for word in words:
for c, count in collections.Counter(word).items():
if count > charsCount[c]:
break
el... | find-words-that-can-be-formed-by-characters | python 3 || counter solution || O(C)/O(1) | dereky4 | 0 | 176 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,942 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/1675247/Python3-dollarolution | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
def dicts(s):
d = {}
for i in s:
if i not in d:
d[i] = 1
else:
d[i] += 1
return d
s, count = 0, 0
d1 = ... | find-words-that-can-be-formed-by-characters | Python3 $olution | AakRay | 0 | 154 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,943 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/1508960/Python-solution-with-explanation-(for-loop-%2B-letter-count) | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
word_char_ct = 0
for word in words:
# initially assume that chars can create word
boolean_ind = True
# iterate thru each letter in word, if a word has MORE letters than available in ... | find-words-that-can-be-formed-by-characters | Python solution with explanation (for loop + letter count) | jjluxton | 0 | 93 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,944 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/1081328/Python3-freq-table | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
fc = {}
for c in chars: fc[c] = 1 + fc.get(c, 0)
ans = 0
for word in words:
fw = {}
for c in word: fw[c] = 1 + fw.get(c, 0)
if all(fw[c] <= fc.get(c, 0) for c... | find-words-that-can-be-formed-by-characters | [Python3] freq table | ye15 | 0 | 76 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,945 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/1081328/Python3-freq-table | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
ans = 0
fc = Counter(chars)
for word in words:
if not Counter(word) - fc: ans += len(word)
return ans | find-words-that-can-be-formed-by-characters | [Python3] freq table | ye15 | 0 | 76 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,946 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/1039312/Python3-easy-solution-using-Counter | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
count = 0
for i in words:
a = Counter(i)
b = Counter(chars)
flag = True
for j in i:
if not b[j] >= a[j]:
flag = False
if fla... | find-words-that-can-be-formed-by-characters | Python3 easy solution using Counter | EklavyaJoshi | 0 | 130 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,947 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/553095/Python-3-Solution-using-dictionary | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
good_words = []
usable_chars_dict = {}
for char in chars:
usable_chars_dict[char] = usable_chars_dict.get(char, 0) + 1
for word in words:
available_chars = usable_chars_dict.copy()
... | find-words-that-can-be-formed-by-characters | Python 3 Solution using dictionary | duyh | 0 | 174 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,948 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/1778478/6-Lines-Python-Solution-oror-Faster-than-96-oror-Memory-less-than-94 | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
ans = 0
for word in words:
for char in word:
if word.count(char) > chars.count(char): break
else: ans += len(word)
return ans | find-words-that-can-be-formed-by-characters | 6 Lines Python Solution || Faster than 96% || Memory less than 94% | Taha-C | -1 | 133 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,949 |
https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/discuss/1722868/Python-60.01-Faster-180ms | class Solution:
def countCharacters(self, words: List[str], chars: str) -> int:
#vars to hold values
wordlen = 0
runningsum = 0
charedits = chars
#evaluate words individually
for word in words:
wordlen = 0
charedits = chars
... | find-words-that-can-be-formed-by-characters | Python 60.01% Faster, 180ms | ovidaure | -1 | 172 | find words that can be formed by characters | 1,160 | 0.677 | Easy | 17,950 |
https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree/discuss/848959/BFS-Python-solution-with-comments! | class Solution:
def maxLevelSum(self, root: TreeNode) -> int:
queue = deque() #init a queue for storing nodes as we traverse the tree
queue.append(root) #first node (level = 1) inserted
#bfs = [] #just for understanding- this will be a bfs list to store nodes as we conduct... | maximum-level-sum-of-a-binary-tree | BFS Python solution - with comments! | tintsTy | 1 | 58 | maximum level sum of a binary tree | 1,161 | 0.661 | Medium | 17,951 |
https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree/discuss/2825715/Python-DFS | class Solution:
def maxLevelSum(self, root: Optional[TreeNode]) -> int:
def dfs(node, level):
if not node:
return
sums[level] += node.val
dfs(node.left, level + 1)
dfs(node.right, level +1 )
sums = ... | maximum-level-sum-of-a-binary-tree | Python, DFS | blue_sky5 | 0 | 2 | maximum level sum of a binary tree | 1,161 | 0.661 | Medium | 17,952 |
https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree/discuss/2791988/Python-solution | class Solution:
def maxLevelSum(self, root: Optional[TreeNode]) -> int:
levels = []
sums = float('-inf')
res = 0
def levelorder(node, level):
if level >= len(levels):
levels.append([])
if node:
levels[level... | maximum-level-sum-of-a-binary-tree | Python solution | maomao1010 | 0 | 5 | maximum level sum of a binary tree | 1,161 | 0.661 | Medium | 17,953 |
https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree/discuss/2431412/python-bfs-beginner-friendly | class Solution:
def maxLevelSum(self, root: Optional[TreeNode]) -> int:
q = collections.deque()
tracker = [float('-inf'),0]
q.append(root)
level = 1
while q:
levelSum = 0
for _ in range(len(q)):
node = q.popleft()
lev... | maximum-level-sum-of-a-binary-tree | python bfs beginner friendly | scr112 | 0 | 17 | maximum level sum of a binary tree | 1,161 | 0.661 | Medium | 17,954 |
https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree/discuss/2209920/Short-and-Very-Intuitive-with-inline-comment | class Solution:
def maxLevelSum(self, root: Optional[TreeNode]) -> int:
if not root:
return None
q=deque()
max_sum=-10**18 #assume the max_level_sum to be -10**18
level=0 #this is current level
level_ans=0 #this will store the level at which the max_sum ha... | maximum-level-sum-of-a-binary-tree | Short and Very Intuitive with inline-comment | Taruncode007 | 0 | 23 | maximum level sum of a binary tree | 1,161 | 0.661 | Medium | 17,955 |
https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree/discuss/1984012/Python-solution-100-Test-Cases | class Solution:
def maxLevelSum(self, root: Optional[TreeNode]) -> int:
def getHeight(root):
if root.left==None and root.right==None:
return 0
left=0
if root.left!=None:
left = getHeight(root.left)
right=0
if root.ri... | maximum-level-sum-of-a-binary-tree | Python solution 100% Test Cases | Siddharth_singh | 0 | 25 | maximum level sum of a binary tree | 1,161 | 0.661 | Medium | 17,956 |
https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree/discuss/1503224/python-dfs | class Solution:
def maxLevelSum(self, root: Optional[TreeNode]) -> int:
stack = [(root, 1)]
h = defaultdict(int)
while stack:
node, depth = stack.pop()
if node:
h[depth] += node.val
stack.append((node.left, 1+depth))
... | maximum-level-sum-of-a-binary-tree | python dfs | byuns9334 | 0 | 95 | maximum level sum of a binary tree | 1,161 | 0.661 | Medium | 17,957 |
https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree/discuss/1422714/PYTHON3-BFS-or-87-Fast! | class Solution:
def maxLevelSum(self, root: Optional[TreeNode]) -> int:
if not root :
return
q = deque()
q.append(root)
res = []
while q :
sum = 0
size = len(q)
for _ in range(size) :
node =... | maximum-level-sum-of-a-binary-tree | PYTHON3 - BFS | 87% Fast! | athrvb | 0 | 45 | maximum level sum of a binary tree | 1,161 | 0.661 | Medium | 17,958 |
https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree/discuss/1081332/Python3-bfs-by-level | class Solution:
def maxLevelSum(self, root: TreeNode) -> int:
ans = level = 0
val = -inf
queue = [root]
while queue:
level += 1
newq = []
tmp = 0
for node in queue:
tmp += node.val
if node.left: newq.ap... | maximum-level-sum-of-a-binary-tree | [Python3] bfs by level | ye15 | 0 | 79 | maximum level sum of a binary tree | 1,161 | 0.661 | Medium | 17,959 |
https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree/discuss/981681/python-3-simple-solution-bfs | class Solution:
def maxLevelSum(self, root: TreeNode) -> int:
if not root:
return None
stack = [root]
values = []
while stack:
queue = []
for _ in range(len(stack)):
node = stack.pop(0)
if node.left:
... | maximum-level-sum-of-a-binary-tree | python 3 simple solution bfs | GiorgosMarga | 0 | 45 | maximum level sum of a binary tree | 1,161 | 0.661 | Medium | 17,960 |
https://leetcode.com/problems/maximum-level-sum-of-a-binary-tree/discuss/361116/Python3-level-order-BFS | class Solution:
def maxLevelSum(self, root: TreeNode) -> int:
from collections import deque
if not root: return []
queue, res = deque([root]), []
while queue:
cur_level, size = [], len(queue)
for i in range(size):
node = queue.popleft(... | maximum-level-sum-of-a-binary-tree | Python3 level order BFS | aj_to_rescue | 0 | 41 | maximum level sum of a binary tree | 1,161 | 0.661 | Medium | 17,961 |
https://leetcode.com/problems/as-far-from-land-as-possible/discuss/1158339/A-general-Explanation-w-Animation | class Solution:
def maxDistance(self, grid: List[List[int]]) -> int:
# The # of rows and # of cols
M, N, result = len(grid), len(grid[0]), -1
# A list of valid points
valid_points = {(i, j) for i in range(M) for j in range(N)}
# A double-ended queue of "land" cells
... | as-far-from-land-as-possible | A general Explanation w/ Animation | dev-josh | 8 | 367 | as far from land as possible | 1,162 | 0.486 | Medium | 17,962 |
https://leetcode.com/problems/as-far-from-land-as-possible/discuss/2258475/PYTHON-or-EXPLAINED-WITH-INTUTION-or-FAST-or-BFS-or-WELL-WRITTEN-or | class Solution:
def maxDistance(self, grid: List[List[int]]) -> int:
n = len(grid)
queue = []
vist = [[False for i in range(n)] for j in range(n)]
for i in range(n):
for j in range(n):
if grid[i][j] == 1:
queue.append((i,j,0))
... | as-far-from-land-as-possible | PYTHON | EXPLAINED WITH INTUTION | FAST | BFS | WELL WRITTEN | | reaper_27 | 4 | 71 | as far from land as possible | 1,162 | 0.486 | Medium | 17,963 |
https://leetcode.com/problems/as-far-from-land-as-possible/discuss/2102946/Python-BFS.-FASTER-THAN-94. | class Solution:
def maxDistance(self, grid: list[list[int]]) -> int:
n = len(grid)
dq = deque((i, j) for i in range(n) for j in range(n) if grid[i][j])
res = 0
while dq:
r0, c0 = dq.popleft()
for dr, dc in ((-1, 0), (1, 0), (0, -1), (0, 1)):
r... | as-far-from-land-as-possible | Python BFS. FASTER THAN 94%. | miguel_v | 4 | 182 | as far from land as possible | 1,162 | 0.486 | Medium | 17,964 |
https://leetcode.com/problems/as-far-from-land-as-possible/discuss/1552850/Easy-Approach-oror-Thought-Process-oror-Well-Explained-and-Coded | class Solution:
def maxDistance(self, grid: List[List[int]]) -> int:
n = len(grid)
q = []
dp = [[-1 for _ in range(n)] for _ in range(n)]
def isvalid(i,j):
if 0<=i<n and 0<=j<n and grid[i][j]==0:
return True
return False
for i in range(n):
for j in ... | as-far-from-land-as-possible | 📌📌 Easy-Approach || Thought Process || Well-Explained and Coded 🐍 | abhi9Rai | 4 | 196 | as far from land as possible | 1,162 | 0.486 | Medium | 17,965 |
https://leetcode.com/problems/as-far-from-land-as-possible/discuss/1081337/Python3-multi-source-bfs | class Solution:
def maxDistance(self, grid: List[List[int]]) -> int:
n = len(grid) # dimension
ans = -1
queue = [(i, j) for i in range(n) for j in range(n) if grid[i][j]]
while queue:
newq = []
for i, j in queue:
for ii, jj in (i-1, ... | as-far-from-land-as-possible | [Python3] multi-source bfs | ye15 | 2 | 94 | as far from land as possible | 1,162 | 0.486 | Medium | 17,966 |
https://leetcode.com/problems/as-far-from-land-as-possible/discuss/1035927/Python-BFS-by-your-senpai | class Solution:
def maxDistance(self, grid: List[List[int]]) -> int:
row, col = len(grid),len(grid[0])
queue = deque([])
water_cell = 0
for x in range(row):
for y in range(col):
if grid[x][y] == 1:
queue.append((x,y))
else:
... | as-far-from-land-as-possible | Python BFS by your senpai | Skywalker5423 | 1 | 156 | as far from land as possible | 1,162 | 0.486 | Medium | 17,967 |
https://leetcode.com/problems/as-far-from-land-as-possible/discuss/2846547/Python-oror-BFS-oror-easy | class Solution:
def maxDistance(self, grid: List[List[int]]) -> int:
n, m = len(grid), len(grid[0])
visited = [[False for _ in range(m)] for _ in range(n)]
max_distance = 0
queue = deque()
for i in range(n):
for j in range(m):
if grid[i][j] == 1:
... | as-far-from-land-as-possible | Python || BFS || easy | dhanu084 | 0 | 1 | as far from land as possible | 1,162 | 0.486 | Medium | 17,968 |
https://leetcode.com/problems/as-far-from-land-as-possible/discuss/2645835/Simple-Multi-Source-BFS-Solution-Python | class Solution:
def maxDistance(self, grid: List[List[int]]) -> int:
n = len(grid)
m = len(grid[0])
delrow = [-1,0,1,0]
delcol = [0,1,0,-1]
queue = []
for i in range(n):
for j in range(m):
if grid[i][j] == 1:
... | as-far-from-land-as-possible | Simple Multi Source BFS Solution - Python | abroln39 | 0 | 32 | as far from land as possible | 1,162 | 0.486 | Medium | 17,969 |
https://leetcode.com/problems/as-far-from-land-as-possible/discuss/2420440/Ad-far-from-land-as-possible-oror-Python3-oror-DP | class Solution:
def maxDistance(self, grid: List[List[int]]) -> int:
dist = [[math.inf] * len(grid[0]) for i in range(0, len(grid))]
land = 0
for i in range(0, len(grid)):
for j in range(0, len(grid[0])):
if(grid[i][j] == 1):
dist[i][j... | as-far-from-land-as-possible | Ad far from land as possible || Python3 || DP | vanshika_2507 | 0 | 33 | as far from land as possible | 1,162 | 0.486 | Medium | 17,970 |
https://leetcode.com/problems/as-far-from-land-as-possible/discuss/1810940/Python-BFS | class Solution:
def maxDistance(self, grid: List[List[int]]) -> int:
rows, cols = len(grid), len(grid[0])
directions = ((0, 1), (0, -1), (-1, 0), (1, 0))
# position class to represent land coordinates
Position = namedtuple('Position', ['row', 'col'])
... | as-far-from-land-as-possible | Python BFS | Rush_P | 0 | 69 | as far from land as possible | 1,162 | 0.486 | Medium | 17,971 |
https://leetcode.com/problems/as-far-from-land-as-possible/discuss/1031999/Easy-to-Read-and-Understand-Python-with-Comments! | class Solution:
def maxDistance(self, grid: List[List[int]]) -> int:
rows = len(grid)
cols = len(grid[0])
q = collections.deque()
zeros = 0
# Get our number of zeros and our 1 starting locations.
for row in range(rows):
for col in range(cols):
... | as-far-from-land-as-possible | Easy to Read and Understand Python with Comments! | Pythagoras_the_3rd | -1 | 98 | as far from land as possible | 1,162 | 0.486 | Medium | 17,972 |
https://leetcode.com/problems/last-substring-in-lexicographical-order/discuss/361321/Solution-in-Python-3-(beats-100) | class Solution:
def lastSubstring(self, s: str) -> str:
S, L, a = [ord(i) for i in s] + [0], len(s), 1
M = max(S)
I = [i for i in range(L) if S[i] == M]
if len(I) == L: return s
while len(I) != 1:
b = [S[i + a] for i in I]
M, a = max(b), a + 1
I = [I[i] for i, j in enumera... | last-substring-in-lexicographical-order | Solution in Python 3 (beats 100%) | junaidmansuri | 3 | 742 | last substring in lexicographical order | 1,163 | 0.35 | Hard | 17,973 |
https://leetcode.com/problems/last-substring-in-lexicographical-order/discuss/2263217/PYTHON-or-EXPLANATION-WITH-PHOTO-or-LINEAR-TIME-or-EASY-or-INTUITIVE-or | class Solution:
def lastSubstring(self, s: str) -> str:
n = len(s)
cmax = max(s)
indexes = [ i for i,c in enumerate(s) if c == cmax ]
gap = 1
while len(indexes) > 1:
new_indexes = []
cmax = max(s[i+gap] for i in indexes if i+gap < n)
for i,... | last-substring-in-lexicographical-order | PYTHON | EXPLANATION WITH PHOTO | LINEAR TIME | EASY | INTUITIVE | | reaper_27 | 1 | 273 | last substring in lexicographical order | 1,163 | 0.35 | Hard | 17,974 |
https://leetcode.com/problems/last-substring-in-lexicographical-order/discuss/1081371/Python3-brute-force | class Solution:
def lastSubstring(self, s: str) -> str:
return max(s[i:] for i in range(len(s))) | last-substring-in-lexicographical-order | [Python3] brute-force | ye15 | 1 | 222 | last substring in lexicographical order | 1,163 | 0.35 | Hard | 17,975 |
https://leetcode.com/problems/last-substring-in-lexicographical-order/discuss/1081371/Python3-brute-force | class Solution:
def lastSubstring(self, s: str) -> str:
ii = k = 0
i = 1
while i + k < len(s):
if s[ii+k] == s[i+k]: k += 1
else:
if s[ii+k] > s[i+k]: i += k+1
else:
ii = max(ii+k+1, i)
i = ii+1... | last-substring-in-lexicographical-order | [Python3] brute-force | ye15 | 1 | 222 | last substring in lexicographical order | 1,163 | 0.35 | Hard | 17,976 |
https://leetcode.com/problems/last-substring-in-lexicographical-order/discuss/482282/Python3-two-methods | class Solution:
def lastSubstring(self, s: str) -> str:
i,j,k,n = 0,1,0,len(s)
while j+k<n:
if s[i+k]==s[j+k]:
k+=1
continue
elif s[i+k]>s[j+k]:
j+=k+1
else:
i=max(j,i+k+1)
j=i+1
k=0
return s[i:]
def lastSubstringBF(self, s: str) -> str:
if len(s)<=1: return s
i,res=1,set()... | last-substring-in-lexicographical-order | Python3 two methods | jb07 | 1 | 361 | last substring in lexicographical order | 1,163 | 0.35 | Hard | 17,977 |
https://leetcode.com/problems/last-substring-in-lexicographical-order/discuss/1073707/Python-O(n)-less-than-10-lines | class Solution:
def lastSubstring(self, s: str) -> str:
max_substring = ""
max_char = ""
for i in range(len(s)):
if s[i] >= max_char:
max_char = s[i]
max_substring = max(max_substring, s[i : ])
return max_substring | last-substring-in-lexicographical-order | Python O(n) less than 10 lines | michaellin986 | -6 | 630 | last substring in lexicographical order | 1,163 | 0.35 | Hard | 17,978 |
https://leetcode.com/problems/invalid-transactions/discuss/670649/Simple-clean-python-only-10-lines | class Solution:
def invalidTransactions(self, transactions: List[str]) -> List[str]:
invalid = []
for i, t1 in enumerate(transactions):
name1, time1, amount1, city1 = t1.split(',')
if int(amount1) > 1000:
invalid.append(t1)
continue
... | invalid-transactions | Simple clean python - only 10 lines | auwdish | 7 | 1,600 | invalid transactions | 1,169 | 0.312 | Medium | 17,979 |
https://leetcode.com/problems/invalid-transactions/discuss/2507495/Python-Solution-or-Hashmap-or-Set | class Solution:
def invalidTransactions(self, transactions: List[str]) -> List[str]:
hashmap = {}
#Hashset is used to skip redudant transactions being added to the result
#We will only store index of the transaction because the same transaction can repeat.
result = set()
for i, t in enum... | invalid-transactions | Python Solution | Hashmap | Set | reeteshz | 5 | 684 | invalid transactions | 1,169 | 0.312 | Medium | 17,980 |
https://leetcode.com/problems/invalid-transactions/discuss/2536635/Python%3A-very-simple-and-straightforward-solution | class Solution:
def invalidTransactions(self, transactions: List[str]) -> List[str]:
cities = defaultdict(lambda: defaultdict(list))
output = []
#build city map.
for t in transactions:
name, time, amount, city = t.split(',')
cities[city][name].append(time)... | invalid-transactions | Python: very simple and straightforward solution | jesse14 | 3 | 1,100 | invalid transactions | 1,169 | 0.312 | Medium | 17,981 |
https://leetcode.com/problems/invalid-transactions/discuss/2827880/Python3-Linear-time-Solution-O(N) | class Solution(object):
def invalidTransactions(self, transactions: List[str]) -> List[str]:
# make a defaultdict to save the transactions first
mapped_transactions = collections.defaultdict(lambda: collections.defaultdict(set))
# go through the transactions and save all of them in... | invalid-transactions | [Python3] - Linear time Solution O(N) | Lucew | 2 | 45 | invalid transactions | 1,169 | 0.312 | Medium | 17,982 |
https://leetcode.com/problems/invalid-transactions/discuss/2794530/Python3-Solution-faster-than-99.90-(with-HashMap) | class Solution:
def invalidTransactions(self, transactions: List[str]) -> List[str]:
invalid = []
txn = collections.defaultdict(list)
for trn in transactions:
name, time, amount, city = trn.split(",")
txn[name].append([time,amount,city])
for ... | invalid-transactions | Python3 Solution faster than 99.90% (with HashMap) | aikyab | 1 | 36 | invalid transactions | 1,169 | 0.312 | Medium | 17,983 |
https://leetcode.com/problems/invalid-transactions/discuss/2739439/Brute-Force-Easy-To-Understand-Python-Solution | class Solution:
'''
transaction:
- name
- time
- city
- amount
transaction is possibly invalid if;
- amount is over 1000
OR all of the following are met;
- within 60 minutes of other_transaction
- same name as another other_transaction
... | invalid-transactions | Brute Force Easy-To-Understand Python Solution | trietostopme | 0 | 17 | invalid transactions | 1,169 | 0.312 | Medium | 17,984 |
https://leetcode.com/problems/invalid-transactions/discuss/2733602/Python-O(n2)-with-HashMap | class Solution:
def invalidTransactions(self, transactions: List[str]) -> List[str]:
nameMap = collections.defaultdict(list)
ans = []
for t in transactions:
splitT = t.split(",")
name = splitT[0]
time = splitT[1]
amount = splitT[2]
... | invalid-transactions | Python O(n^2) with HashMap | chinclashs | 0 | 40 | invalid transactions | 1,169 | 0.312 | Medium | 17,985 |
https://leetcode.com/problems/invalid-transactions/discuss/2263406/or-PYTHON-WELL-WRITTEN-or-O(N2)-or-EASY-or-FAST-or | class Solution:
def invalidTransactions(self, transactions: List[str]) -> List[str]:
n = len(transactions)
for i in range(n):
transactions[i] = transactions[i].split(',')
transactions.sort(key = lambda x: int(x[1]))
ans = []
failed = [False for i in range(n)]
... | invalid-transactions | | PYTHON WELL WRITTEN | O(N^2) | EASY | FAST | | reaper_27 | 0 | 204 | invalid transactions | 1,169 | 0.312 | Medium | 17,986 |
https://leetcode.com/problems/invalid-transactions/discuss/1565057/Python-O(N-%2B-KlogK)-Time-Straightforward-Python-Solutions-beats-100 | class Solution:
def invalidTransactions(self, transactions: List[str]) -> List[str]:
N = len(transactions)
# parse strings in transactions for later references
data = [(n, int(t), int(m), c) for n, t, m, c in map(lambda s: s.split(','), transactions)]
get_name = lambda i: data[i][... | invalid-transactions | [Python] O(N + KlogK) Time Straightforward Python Solutions, beats 100% | licpotis | 0 | 238 | invalid transactions | 1,169 | 0.312 | Medium | 17,987 |
https://leetcode.com/problems/invalid-transactions/discuss/1084468/Python3-brute-force | class Solution:
def invalidTransactions(self, transactions: List[str]) -> List[str]:
flag = [False]*len(transactions)
for i, transaction in enumerate(transactions):
n, t, a, c = transaction.split(",")
if int(a) > 1000: flag[i] = True
for ii in range(i+1, len(tran... | invalid-transactions | [Python3] brute-force | ye15 | 0 | 142 | invalid transactions | 1,169 | 0.312 | Medium | 17,988 |
https://leetcode.com/problems/invalid-transactions/discuss/366512/Solution-in-Python-3 | class Solution:
def invalidTransactions(self, T: List[str]) -> List[str]:
u, V = list(map(lambda x: [x[0],x[3],int(x[1]),int(x[2])], map(lambda x: x.split(','), T))), set()
N = {i[0]:[] for i in u}
for i,[n,c,t,a] in enumerate(u):
N[n].append([t,c,i])
if a > 1000: V.add(T[i])
for i i... | invalid-transactions | Solution in Python 3 | junaidmansuri | 0 | 750 | invalid transactions | 1,169 | 0.312 | Medium | 17,989 |
https://leetcode.com/problems/invalid-transactions/discuss/366522/Simple-13-Line-Python-Solution | class Solution:
def invalidTransactions(self, ts: List[str]) -> List[str]:
nts = [t.split(',') for t in ts]
nts = sorted([[a, int(b), int(c), d] for a, b, c, d in nts])
res = set()
for a in nts:
if a[2] > 1000: res.add(','.join(map(str,a)))
for i in range(len(nts)... | invalid-transactions | Simple 13 Line Python Solution | code_report | -1 | 785 | invalid transactions | 1,169 | 0.312 | Medium | 17,990 |
https://leetcode.com/problems/compare-strings-by-frequency-of-the-smallest-character/discuss/401039/Python-Simple-Code-Memory-efficient | class Solution:
def numSmallerByFrequency(self, queries: List[str], words: List[str]) -> List[int]:
def f(s):
t = sorted(list(s))[0]
return s.count(t)
query = [f(x) for x in queries]
word = [f(x) for x in words]
m = []
for x in query:
count = 0
for y in word:
if y>x:
count+=1
m.append... | compare-strings-by-frequency-of-the-smallest-character | Python Simple Code Memory efficient | saffi | 14 | 1,700 | compare strings by frequency of the smallest character | 1,170 | 0.614 | Medium | 17,991 |
https://leetcode.com/problems/compare-strings-by-frequency-of-the-smallest-character/discuss/1236328/Python3-Brute-Force-Solution | class Solution:
def numSmallerByFrequency(self, queries: List[str], words: List[str]) -> List[int]:
def f(x):
return x.count(min(x))
ans = []
for i in queries:
count = 0
for j in words:
if(f(i) < f(j)):
count +... | compare-strings-by-frequency-of-the-smallest-character | [Python3] Brute Force Solution | VoidCupboard | 3 | 77 | compare strings by frequency of the smallest character | 1,170 | 0.614 | Medium | 17,992 |
https://leetcode.com/problems/compare-strings-by-frequency-of-the-smallest-character/discuss/1203192/Python-Easy-Solution | class Solution:
def numSmallerByFrequency(self, queries: List[str], words: List[str]) -> List[int]:
q = []
for query in queries:
query = sorted(query)
temp = query.count(query[0])
q.append(temp)
w = []
for word in words:
word = sorted(... | compare-strings-by-frequency-of-the-smallest-character | Python Easy Solution | iamkshitij77 | 2 | 97 | compare strings by frequency of the smallest character | 1,170 | 0.614 | Medium | 17,993 |
https://leetcode.com/problems/compare-strings-by-frequency-of-the-smallest-character/discuss/2558508/Python3-or-Solved-Using-Binary-Search-By-Translating-Each-and-Every-word-into-Function-Value | class Solution:
#Let n = len(queries) and m = len(words)
#Time-Complexity: O(m + mlog(m) + n*log(m)) -> O(mlog(m) + nlog(m))
#Space-Complexity: O(10*m + n*10 + m) -> O(n + m)
def numSmallerByFrequency(self, queries: List[str], words: List[str]) -> List[int]:
#Approach: Traverse linearly... | compare-strings-by-frequency-of-the-smallest-character | Python3 | Solved Using Binary Search By Translating Each and Every word into Function Value | JOON1234 | 1 | 20 | compare strings by frequency of the smallest character | 1,170 | 0.614 | Medium | 17,994 |
https://leetcode.com/problems/compare-strings-by-frequency-of-the-smallest-character/discuss/416774/Python3 | class Solution:
def numSmallerByFrequency(self, queries: List[str], words: List[str]) -> List[int]:
queries_frequecy = [self.f(i) for i in queries]
words_frequecy = [self.f(i) for i in words]
words_frequecy.sort()
res = []
length = len(words_frequecy)
for i in range(l... | compare-strings-by-frequency-of-the-smallest-character | [Python3] | zhanweiting | 1 | 235 | compare strings by frequency of the smallest character | 1,170 | 0.614 | Medium | 17,995 |
https://leetcode.com/problems/compare-strings-by-frequency-of-the-smallest-character/discuss/2768919/python3-count-and-binary-search-sol-for-reference. | class Solution:
def smallestCharFreq(self, s):
sc = s[0]
cnt = 1
for idx in range(1,len(s)):
c = s[idx]
if c < sc:
cnt = 1
sc= c
elif c == sc:
cnt += 1
return cnt
def numSmallerByFrequency(sel... | compare-strings-by-frequency-of-the-smallest-character | [python3] count and binary search sol for reference. | vadhri_venkat | 0 | 7 | compare strings by frequency of the smallest character | 1,170 | 0.614 | Medium | 17,996 |
https://leetcode.com/problems/compare-strings-by-frequency-of-the-smallest-character/discuss/2498001/easy-python-solution | class Solution:
def numSmallerByFrequency(self, queries: List[str], words: List[str]) -> List[int]:
output, query_count, word_count = [], [], []
for word in queries :
query_word = [ch for ch in word]
query_word.sort()
f_query_word = query_word.count(query_word[0]... | compare-strings-by-frequency-of-the-smallest-character | easy python solution | sghorai | 0 | 14 | compare strings by frequency of the smallest character | 1,170 | 0.614 | Medium | 17,997 |
https://leetcode.com/problems/compare-strings-by-frequency-of-the-smallest-character/discuss/2485310/Beginner-friendly-python-solution | class Solution:
def numSmallerByFrequency(self, queries: List[str], words: List[str]) -> List[int]:
res = []
words = [w.count(min(w)) for w in words]
queries = [q.count(min(q)) for q in queries]
for q in queries:
count = 0
for w in words:
count... | compare-strings-by-frequency-of-the-smallest-character | Beginner friendly python solution | yhc22593 | 0 | 3 | compare strings by frequency of the smallest character | 1,170 | 0.614 | Medium | 17,998 |
https://leetcode.com/problems/compare-strings-by-frequency-of-the-smallest-character/discuss/2263490/PYTHON-or-AS-INTERVIEWER-WANTS-or-WITHOUT-USING-PRE-DEFINED-FUNCTIONS | class Solution:
def numSmallerByFrequency(self, queries: List[str], words: List[str]) -> List[int]:
def freq(word):
l = len(word)
ans,minn = 1,word[0]
for i in range(1,l):
if word[i] == minn: ans += 1
elif word[i] < minn :
... | compare-strings-by-frequency-of-the-smallest-character | PYTHON | AS INTERVIEWER WANTS | WITHOUT USING PRE DEFINED FUNCTIONS | reaper_27 | 0 | 36 | compare strings by frequency of the smallest character | 1,170 | 0.614 | Medium | 17,999 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.